PPR Maths nbk MODULE 12 SKIM TUISYEN FELDA (STF) MATEMATIK SPM “ENRICHMENT” TOPIC : MATRICES TIME : 2 HOURS
1.
(a)
⎛3 − 2⎞ ⎟⎟ is m ⎝5 − 4⎠
The inverse matrix of ⎜⎜
⎛− 4 ⎜⎜ ⎝− 5
n⎞ ⎟ 3 ⎟⎠
Find the value of m and of n. (b)
Answer : (a)
(b)
Hence, using matrices, solve the following simultaneous equations : 3x – 2y = 8 5x – 4y = 13
PPR Maths nbk 2.
(a)
⎛ m 3⎞ ⎟⎟ 2 n ⎝ ⎠
Given that G = ⎜ ⎜
and the inverse matrix of G is
1 ⎛ 4 − 3⎞ ⎜ ⎟, 14 ⎜⎝ − 2 m ⎟⎠
find the value of m and of n. (b)
Hence, using matrices, calculate the value of p and of q that satisfies the following equation :
⎛ p⎞ ⎛ 1 ⎞ G⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟ ⎝ q ⎠ ⎝ − 8⎠ Answer : (a)
(b)
PPR Maths nbk 3.
⎛ −1 2⎞ ⎛1 0⎞ ⎟⎟ = ⎜⎜ ⎟⎟, A⎜⎜ − 3 5 0 1 ⎝ ⎠ ⎝ ⎠
(a)
Given that
(b)
Hence, using the matrix method, find the value of r and s which satisfy the simultaneous equations below. -r + 2s = -4 -3r + 5s = -9
Answer : (a)
(b)
find matrix A.
PPR Maths nbk 4.
⎛ 4 5⎞ ⎟⎟ and matrix PQ = 6 8 ⎝ ⎠
Given matrix P = ⎜ ⎜ (a) (b)
Answer : (a)
(b)
⎛1 0⎞ ⎜⎜ ⎟⎟ 0 1 ⎝ ⎠
Find the matrix Q. Hence, calculate by using the matrix method, the values of m and n that satisfy the following simultaneous linear equations : 4m + 5n = 7 6m + 8n = 10
PPR Maths nbk 5.
⎛ 4 − 3⎞ ⎟⎟ , 8 − 5 ⎝ ⎠
Given the matrix P is ⎜ ⎜
⎛1 0⎞ ⎟⎟ 0 1 ⎝ ⎠
(a)
Find the matrix Q so that PQ = ⎜⎜
(b)
Hence, calculate the values of h and k, which satisfy the matrix equation:
⎛ 4 − 3 ⎞⎛ h ⎞ ⎛ − 7 ⎞ ⎜⎜ ⎟⎟⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟ 8 − 5 k − 11 ⎝ ⎠⎝ ⎠ ⎝ ⎠ Answer : (a)
(b)
PPR Maths nbk 6.
⎛ k 6⎞ ⎟⎟, find the value of k if matrix M has no inverse. − 4 2 ⎝ ⎠
(a)
Given matrix M = ⎜ ⎜
(b)
Given the matrix equations
⎛ 7 − 6 ⎞⎛ x ⎞ ⎛ − 4 ⎞ ⎜⎜ ⎟⎟⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟ ⎝ − 5 8 ⎠⎝ y ⎠ ⎝ 1 ⎠ (i) (ii) Answer : (a)
(b)
⎛ x⎞
1 ⎛ 8 6 ⎞⎛ − 4 ⎞
and ⎜ ⎟ = ⎜ ⎜ y ⎟ h ⎜ 5 7 ⎟⎟⎜⎜ 1 ⎟⎟ ⎝ ⎠ ⎝ ⎠⎝ ⎠
Find the value of h Hence, find the value of x and y.
PPR Maths nbk 7.
⎛2 5 ⎞ ⎟⎟ does not have an inverse matrix. k − 2 ⎝ ⎠
It is given that matrix P = ⎜ ⎜ (a) (b)
Find the value of k. If k = 1, find the inverse matrix of P and hence, using matrices, find the values of x and y that satisfy the following simultaneous linear equations. 2x + 5y = 13 x - 2y = -7
Answer : (a)
(b)
PPR Maths nbk 8.
⎛ 2 4⎞ ⎛ 2 4⎞ ⎟⎟ M = ⎜⎜ ⎟⎟ 1 3 1 3 ⎝ ⎠ ⎝ ⎠
(a)
Find matrix M such that ⎜ ⎜
(b)
Using matrices, calculate the values of x and y that satisfy the following matrix equation.
⎛ 2 4 ⎞⎛ x ⎞ ⎛ 6 ⎞ ⎜⎜ ⎟⎟⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟ ⎝ 1 3 ⎠⎝ y ⎠ ⎝ 5 ⎠
Answer : (a)
(b)
PPR Maths nbk
9.
⎛3 −1⎞ ⎟⎟ . 5 − 2 ⎝ ⎠
(a)
Find the inverse of matrix ⎜ ⎜
(b)
Hence, using matrices, calculate the values of d and e that satisfy the following simultaneous equations : 2d – e = 7 5d – e = 16
Answer : (a)
(b)
PPR Maths nbk
10.
⎛1 − 2⎞ ⎟⎟ , find 2 5 ⎝ ⎠
Given matrix M = ⎜ ⎜ (a) (b)
the inverse matrix of M hence, using matrices, the values of u and v that satisfy the following simultaneous equations : u – 2v = 8 2u + 5v = 7
Answer : (a)
(b)
PPR Maths nbk MODULE 12 - ANSWERS TOPIC : MATRICES
1.
m= −
(a)
1 2
1m
n =2 (b) ⎛ 3
⎜⎜ ⎝5
1m
− 2⎞ ⎛ x ⎞ ⎛8 ⎞ ⎟⎜ ⎟ = ⎜ ⎟ − 4 ⎟⎠ ⎜⎝ y ⎟⎠ ⎜⎝13 ⎟⎠ ⎛ x ⎞ 1 ⎛− 4 ⎜⎜ ⎟⎟ = ⎜⎜ ⎝ y⎠ 2 ⎝− 5 x=3 y= −
2.
(a)
⎜⎜ ⎝2
2 ⎞⎛ 8 ⎞ ⎟⎜ ⎟ 3 ⎟⎠⎜⎝13 ⎟⎠
1 2
1m
1m 1m
3 ⎞⎛ p ⎞ ⎟⎜ ⎟ = 4 ⎟⎠⎜⎝ q ⎟⎠
⎛ 1 ⎞ ⎜⎜ ⎟⎟ ⎝ − 8⎠
⎛ p ⎞ 1 ⎛ 4 − 3 ⎞⎛ 1 ⎞ ⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟⎜⎜ ⎟⎟ ⎝ q ⎠ 14 ⎝ − 2 5 ⎠⎝ − 8 ⎠ p=2 q = -3
3.
(a)
(b)
4.
(a)
⎛5 ⎜3 ⎝
A=⎜
1m
1m
n =4 m=5
(b) ⎛ 5
1m
1m
1m
1m 1m
− 2⎞ ⎟ − 1 ⎟⎠
2m
⎛ −1 2⎞ ⎛ r ⎞ ⎛ − 4⎞ ⎜⎜ ⎟⎟ ⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟ ⎝− 3 5⎠ ⎝ s ⎠ ⎝ − 9⎠
1m
⎛ r ⎞ 1 ⎛ 5 − 2 ⎞⎛ − 4 ⎞ ⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟⎜⎜ ⎟⎟ ⎝ s ⎠ 1 ⎝ 3 − 1 ⎠⎝ − 9 ⎠
1m
P=
r = -2
1m
s = -3
1m 1m
1 ⎛ 8 − 5⎞ ⎜ ⎟ 32 − 30 ⎜⎝ − 6 4 ⎟⎠
PPR Maths nbk
(b)
5.
(a)
(b)
6.
1 ⎛ 8 − 5⎞ ⎜ ⎟ = 2 ⎜⎝ − 6 4 ⎟⎠
1m
⎛ 4 5 ⎞⎛ m ⎞ ⎛ 7 ⎞ ⎜⎜ ⎟⎟⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟ ⎝ 6 8 ⎠⎝ n ⎠ ⎝10 ⎠
1m
⎛ m ⎞ 1 ⎛ 8 − 5 ⎞⎛ 7 ⎞ ⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟⎜⎜ ⎟⎟ ⎝ n ⎠ 2 ⎝ − 6 4 ⎠⎝10 ⎠
1m
m=3
1m
n = -1
1m
⎛ − 5 3 ⎞ 1m 1 ⎜ ⎟ − 20 − (−24) ⎜⎝ 8 4 ⎟⎠ 1 ⎛ − 5 3⎞ 1m ⎟ = ⎜⎜ 4 ⎝ 8 4 ⎟⎠
P =
⎛ 4 − 3 ⎞⎛ h ⎞ ⎛ − 7 ⎞ ⎜⎜ ⎟⎟⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟ ⎝ 8 − 5 ⎠⎝ k ⎠ ⎝ − 11⎠ ⎛ h ⎞ 1 ⎛ − 5 3 ⎞⎛ − 7 ⎞ ⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟⎜⎜ ⎟⎟ ⎝ k ⎠ 2 ⎝ 8 4 ⎠⎝ − 11⎠ 1⎛ 2 ⎞ ⎟ = ⎜⎜ 2 ⎝ − 100 ⎟⎠
1m
1m
h=1 k = -50
1m 1m
(a)
k = -12
1m
(b)
(i)
h = 26
⎛ x⎞ 1 ⎛ 8 6 ⎞⎛ − 4 ⎞ ⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟⎜⎜ ⎟⎟ ⎝ y ⎠ 26 ⎝ 5 7 ⎠⎝ 1 ⎠ 1 ⎛ − 26 ⎞ ⎜ ⎟ = 26 ⎜⎝ − 13 ⎟⎠
1m
PPR Maths nbk (ii)
1m
1m
x = -1 y= −
7.
(a)
1m
1 2
- 4 – 5k = 0
1m
1m
5k = -4 k= − (b)
8.
4 5
1m
⎛ 2 5 ⎞⎛ x ⎞ ⎛ 13 ⎞ ⎜⎜ ⎟⎟⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟ ⎝ 1 − 2 ⎠⎝ y ⎠ ⎝ − 7 ⎠
1m
⎛ x⎞ 1 ⎛ − 2 − 5 ⎞⎛ 13 ⎞ ⎜⎜ ⎟⎟ = − ⎜⎜ ⎟⎟⎜⎜ ⎟⎟ 1 2 − y 9 ⎝ ⎠ ⎝ ⎠⎝ − 7 ⎠
1m
x = -1 y=3
1m 1m
⎛1 0⎞ ⎟⎟ ⎝0 1⎠
(a)
M= ⎜ ⎜
2m
(b)
⎛ x⎞ 1 ⎛ 3 − 4 ⎞⎛ 6 ⎞ ⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟⎜⎜ ⎟⎟ ⎝ y ⎠ 6 − 4 ⎝ − 1 2 ⎠⎝ 5 ⎠
1m
1 ⎛ 3 − 4 ⎞⎛ 6 ⎞ ⎜ ⎟⎜ ⎟ 2 ⎜⎝ − 1 2 ⎟⎠⎜⎝ 5 ⎟⎠ 1 ⎛ − 2⎞ = ⎜⎜ ⎟⎟ 2⎝ 4 ⎠
=
x = -1 y=2
1m
1m 1m
PPR Maths nbk 9.
(a)
1 ⎛ − 2 1⎞ ⎜ ⎟ − 6 + 5 ⎜⎝ − 5 3 ⎟⎠
1m
1 ⎛ − 2 1⎞ ⎜ ⎟ − 1 ⎜⎝ − 5 3 ⎟⎠
1m
=
− 1 ⎞⎛ d ⎞ ⎛ 7 ⎞ ⎜⎜ ⎟⎟⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟ ⎝ 5 − 3 ⎠⎝ e ⎠ ⎝16 ⎠ ⎛ d ⎞ 1 ⎛ − 3 1 ⎞⎛ 7 ⎞ ⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟⎜⎜ ⎟⎟ ⎝ e ⎠ − 1 ⎝ − 5 2 ⎠⎝16 ⎠
(b) ⎛ 2
=
1m
1m
1 ⎛ − 5⎞ ⎜ ⎟ − 1 ⎜⎝ − 3 ⎟⎠
⎛ 5⎞ = ⎜⎜ ⎟⎟ ⎝ 3⎠ d=5 e=3
10.
(a)
⎛ 5 2⎞ 1 ⎜ ⎟ 5 − (−4) ⎜⎝ − 2 1 ⎟⎠ 1 ⎛ 5 2⎞ ⎟ = ⎜⎜ 9 ⎝ − 2 1 ⎟⎠
− 2 ⎞⎛ u ⎞ ⎛ 8 ⎞ ⎜ 2 5 ⎟⎟⎜⎜ v ⎟⎟ = ⎜⎜ 7 ⎟⎟ ⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎛ u ⎞ 1 ⎛ 5 2 ⎞⎛ 8 ⎞ ⎟⎟⎜⎜ ⎟⎟ ⎜⎜ ⎟⎟ = ⎜⎜ ⎝ v ⎠ 9 ⎝ − 2 1 ⎠⎝ 7 ⎠
(b) ⎛⎜ 1
1m 1m
1m
1m
1m
1m
1 ⎛ 54 ⎞ = ⎜⎜ ⎟⎟ 9 ⎝ − 9⎠ ⎛6⎞ = ⎜⎜ ⎟⎟ ⎝ − 1⎠ u=6 v = −1
1m 1m